\(\left|x-1\right|=2x+3\)
Với \(x-1\ge0\Leftrightarrow x\ge1\)
\(\Leftrightarrow x-1=2x+3\)
\(\Leftrightarrow-x=4\)
\(\Leftrightarrow x=-4\left(L\right)\)
Với \(x-1< 0\Leftrightarrow x< 1\)
\(\Leftrightarrow-x+1=2x+3\)
\(\Leftrightarrow-3x=2\)
\(\Leftrightarrow x=-\dfrac{2}{3}\left(N\right)\)
Vậy \(S=\left\{-\dfrac{2}{3}\right\}\)